NDA II 2019 (Previous Year Question Paper Maths): Solution

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NDA II 2019 (Previous Year Question Paper Maths): Solution 1. Ans. D. We know that the binary number have only two digits ( 0 & 1) As given in the question c > d i.e, c = 1 & d = 0 Then, the given number will be (cdccddcccddd)2 = ( 101100111000)2

Now, add up the higher bit only : 2048 + 512 + 256 + 32 + 16 + 8 = 2872 Hence, option D is correct. 2. Ans. D. Maximum number of subset = 2n ( where n is the number of element in set S) Hence, the maximum subset S have 210 = 1024 Option D is correct. 3. Ans. C. Given equation 4(x - p) (x - q)- r2 = 0, where p, q and r are real numbers Expand it: 4x2 ? 4qx ?4px +4pq ? r2 = 0 It will became : 4x2 ?4(p + q) x + 4pq ? r2 = 0 Now, check the discriminant ( D) D = 16[-(p +q)]2 ? 16(4pq ? r2) D = 16[p2+ q2 + 2pq ? 4pq + r2] D = 16[p2+ q2 - 2pq + r2] D = 16[(p ? q)2 + r2]

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If p, q, r are real number then D > 0 ; hence roots will always be real If p = q & r = 0 then D =0 Hence, both statement are correct. Option C is correct. 4. Ans. A. The given equation is: We can open the mode as

When you calculate the roots of the equations roots are -2, -1 (which is not possible as x > 0 ) roots are 2 , 1 ( which is also not possible as x < 0)

Hence, option A is correct. 5. Ans. C.

Given

Then,

But X should also be greater than 0 (given in the question) 2

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Then Option C is correct. 6. Ans. C.

x 4, y 0 it can be written as

x - 4, y 0 it can also be written as then the common part between both the condition is x -4 and y = 0 hence, option C is correct. 7. Ans. A. Given 3rd, 8th and 13th terms of a GP are p, q and r respectively and nth term of a G.P = arn-1 3rd term = ar2 = p 8th term = ar7 = q 13th term = ar12 = r Then, from above we can clearly see that : (8th term)2 = (3rd term)(13th term) i.e, q2 = pr option A is correct. 8. Ans. C. General terms of an A.P: a, a +d, a + 2d, a + 3d, a + 4d................. Now, S2n = 3n + 14n2 {given} Then, S2 = 3 x 1 + 14 x 12 = 17 S4 = 3x2 + 14x22 = 6 + 56 = 62 S2 = a + a + d = 2a + d = 17 i.e, 2a + d = 17 ......1st eq.

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S4 = a + a + d + a + 2d + a + 3d = 4a + 6d = 62 i.e, 4a + 6d = 62 .......2nd eq. solve 1st and 2nd , we get d = 7 option C is correct. 9. Ans. B. Two digits numbers are {10,11,12,13,14,15,....................99} The first number that is divisible by 4 = 12 And the last number that is divisible by 4 = 96 Number which are divisible by 4 are 12, 16,20,24,28........96 (it is an A.P) the common difference is 4. Then, 96 = a + (n-1)d 96 = 12 + (n ? 1)4 96-12=(n-1)4 Solve this , you will get n = 22 Option B is correct. 10. Ans. D. Given a, b, c be in AP and k 0 be a real number We know that if we multiply divide or subtract an A.P with a constant term then the obtained value will also be form an A.P(properties of an A.P) Hence, all the tree statements are correct. Option D is correct. 11. Ans. C. We have the property : nCr + nCr+1 = n+1Cr+1

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Now, the given expression is : C(47, 4) + C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) + C(47, 3) We can rearrange it as: [C(47, 4) + C(47, 3)]+ C(51, 3) + C(50, 3) + C(49, 3) + C(48, 3) Use above property; [C(48,4) + C(48, 3)]+ C(51, 3) + C(50, 3) + C(49, 3) [C(49,4) + + C(49, 3)] + C(51, 3) + C(50, 3) [C(50,4) + C(50, 3)] + C(51, 3) C(51,4) + C(51,3) C(52,4) Hence, option C is correct. 12. Ans. B.

Given expression :

Constant term = 405 Then,the general term will be = 10Cr(x1/2)10 ? r(kx-2)r Now, we can write it as Kr10Cr x5-r/2x-2r = kr10CrX5-5r/2 For constant term 5 ? 5r/2 = 0 r = 2, then k2 10C2 = 405

Option B is correct. 13. Ans. B.

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Given AUB =

.

From the given condition, we have that (x-a)(x-b)>0

It is given that a < b

It is possible only when

Then, only option B will satisfy the result

Hence, option B is correct.

14. Ans. B.

If n! has 17 zeros, then we have to find the value of n

For the occurance of zero in the result of any multiplication it should be that there is either a multiple of 10 or the multiplication of (2 X 5).

For ex: 10! = 10x9x8x7x6x5x4x3x2x1 (the result will surely consist two zeroes)

In the same ways: 20! Will consist four zeroes.

Hence, option B is correct. 15. Ans. C. For the multiplication of the two matrix the necessary condition is that the column of the Ist matrix will be equal to the raw of the IInd matrix. Here, the order of the matrix A = 3 X 2

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And order of the matrix B = 2 X 2 Then, according to the condition of the matrix only AB will exist but BA will not exist. Option C is correct. 16. Ans. A.

Given expression is :

Here the total number of term in the given expression, when we expand it will be 2n + 1;

Then, the middle term will be n + 1

Hence, (n + 1)th term of the given expression can be written as 2nCn(x2)n(x1)n = 184756x10 Then, 2nCnxn = 184756x10

From above, we can clearly see that n = 10

Option A is correct.

17. Ans. C. Given equation is : (1 + 2x + x2)5 + (1 + 4y + 4y2)5 We have to find the number of terms in the given expression We can write the above equation as : (1 + x)10 + ( 1+ 2y)10 The number of term in (1 + x)10 will be 11 & the number of terms in ( 1+ 2y)10 will also be 11 But a term will be common in both (1 + x)10 & ( 1+ 2y)10 (which is independent of the coefficient of x & y) Then, total number of terms will be 21. Option C is correct. 18. Ans. C. If P(n, r) = 2520 and C(n, r) = 21, then we have to calculate the value of C(n + 1, r + 1)

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We know that nCr =

Then r! =

=120

r! = 120 then, r = 5

Now, we have to apply hit and trail method to calculate the value of n n r n 6 Now, it is given that nC5 = 21 Then, n = 7

Option C is correct. 19. Ans. B.

Given expression is :

Assume Then, we can write it as : x = 2 + Now, after solving, we get x2 ? 2x -1 = 0 x = 1 will be the two roots

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